If $A$ is a $n \times n$ matrix, taking $x$ to be an eigenvector associated with the largest eigenvalue $\lambda_\max$ yields $\sup_{\Vert x \rVert = 1} \langle A x, x \rangle \ge \lambda_\max$.
It is also easy to see that $\sup_{\Vert x \rVert = 1} \langle A x, x \rangle \le \sup_{\Vert x \rVert = 1} \lVert A x\rVert \le \sqrt{\lambda^*_\max}$ where $\lambda^*_\max$ is the largest eigenvalue of $A^*A$.
But is it true that $\sup_{\Vert x \rVert = 1} \langle A x, x \rangle = \lambda_\max$ ?